Question

The general solution of the differential equation $\frac{dy}{dx} + \frac{y^2+y+1}{x^2+x+1} = 0$ is

• A. $x + y + 1 = c (1 + x + y + 2xy)$
• B. $x + y + 1 = c (2 + x + y + 2xy)$
• C. $x + y + 1 = c (1 - x - y - 2xy)$
• D. $x + y + 2 = c (2 - x - y - 2xy)$

Hint:

When combining inverse tangents in differential equations with symmetric forms like $\tan^{-1} U + \tan^{-1} V = C$, the denominator of the argument inside the compound tangent function always follows a $(1 - UV)$ format. Since $U \propto (2x+1)$ and $V \propto (2y+1)$, their product introduces a negative cross-term $-4xy$. This guarantees that the right side of your final equation must contain subtraction terms, isolating option (C).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given a first-order non-linear differential equation. We need to find its general algebraic solution by separating variables.

Step 2: Key Formula or Approach:
1. Rearrange the variables so that $y$ terms are on the left and $x$ terms are on the right: $$\frac{1}{y^2+y+1} dy = -\frac{1}{x^2+x+1} dx$$ 2. Integrate both sides using the standard algebraic integral formula: $$\int \frac{1}{u^2+a^2} du = \frac{1}{a} \tan^{-1}\left(\frac{u}{a}\right)$$ 3. Combine the resulting inverse tangent functions using the identity: $$\tan^{-1} A + \tan^{-1} B = \tan^{-1}\left(\frac{A+B}{1-AB}\right)$$

Step 3: Detailed Explanation:
4. Separate variables and complete the squares for the quadratic expressions in the denominators: $$x^2 + x + 1 = \left(x+\frac{1}{2}\right)^2 + \frac{3}{4} = \left(x+\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2$$ Similarly for $y$, our equation becomes: $$\int \frac{dy}{\left(y+\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = -\int \frac{dx}{\left(x+\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2}$$ 5. Integrate both sides: $$\frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{y+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right) = -\frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{x+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right) + C_1$$ $$\tan^{-1}\left(\frac{2y+1}{\sqrt{3}}\right) + \tan^{-1}\left(\frac{2x+1}{\sqrt{3}}\right) = C_2 \quad \left[\text{where } C_2 = \frac{\sqrt{3}}{2}C_1\right]$$ 6. Apply the tangent addition identity: $$\tan^{-1}\left( \frac{\frac{2y+1}{\sqrt{3}} + \frac{2x+1}{\sqrt{3}}}{1 - \left(\frac{2y+1}{\sqrt{3}}\right)\left(\frac{2x+1}{\sqrt{3}}\right)} \right) = C_2$$ Take the tangent of both sides to remove the inverse operator: $$\frac{\frac{2x+2y+2}{\sqrt{3}}}{\frac{3 - (4xy+2x+2y+1)}{3}} = \tan C_2$$ $$\frac{\sqrt{3}(2x+2y+2)}{2 - 2x - 2y - 4xy} = \tan C_2$$ $$\frac{x+y+1}{1 - x - y - 2xy} = \frac{\tan C_2}{\sqrt{3}} = c$$ 7. Cross-multiply to yield the final form: $$x+y+1 = c(1 - x - y - 2xy)$$

Step 4: Final Answer:
The general solution matches option (C).